Problem: $f(t) = 4t-5+4(h(t))$ $h(n) = 2n-2$ $ f(h(-1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(-1)$ . Then we'll know what to plug into the outer function. $h(-1) = (2)(-1)-2$ $h(-1) = -4$ Now we know that $h(-1) = -4$ . Let's solve for $f(h(-1))$ , which is $f(-4)$ $f(-4) = (4)(-4)-5+4(h(-4))$ To solve for the value of $f$ , we need to solve for the value of $h(-4)$ $h(-4) = (2)(-4)-2$ $h(-4) = -10$ That means $f(-4) = (4)(-4)-5+(4)(-10)$ $f(-4) = -61$